Question: $\int^{\pi}_{0}\sec^2\left(\dfrac{x}{3}\right)\,dx\, = $
Explanation: Strategy Let's first find the indefinite integral $\int\sec^2\left(\dfrac{x}{3}\right)\,dx\, $. Then we can use the result to compute the definite integral. Finding the indefinite integral To find $\int\sec^2\left(\dfrac{x}{3}\right)\,dx\, $ we can use U-substitution. If we let $ {u=\dfrac{x}{3}}$, then ${du=\dfrac13 \, dx}$ and ${ \,dx=3\, du}$. So we have: $\begin{aligned}\int\sec^2\left({\dfrac{x}{3}}\right)\,{dx}\, &=\int\sec^2( u)\,{3\, du}\,\\\\\\\\ &=3\int\sec^2(u)\,du\\\\\\\\ &=3\tan(u)+C\\\\\\\\ &=3\tan\left(\dfrac{x}{3}\right)+C \end{aligned}$ Evaluating the definite integral Now let's compute the definite integral: $\begin{aligned}\int^{\pi}_{0}\sec^2\left(\dfrac{x}{3}\right)\,dx\, &= 3\tan\left(\dfrac{x}{3}\right)\Bigg|^{\pi}_0\\\\\\\\ &=3\left(\tan\left(\dfrac{\pi}{3}\right)-\tan(0)\right)\\\\\\\\ &=3(\sqrt{3}-0)\\\\\\\\ &=3\sqrt{3}\end{aligned}$ [Did we have to find the indefinite integral first?] The answer $\int^{\pi}_{0}\sec^2\left(\dfrac{x}{3}\right)\,dx\, = 3\sqrt{3}$